Orderable groups with applications to topology Dale
... Proposition: Suppose p, q are non-negative and r, s are nonpositive integers (or vice-versa). Then π1(Mϕ) is not left-orderable. Proof: The group π1(Mϕ) has presentation with generators a, b, x, y and relations: a2 = b2, x2 = y 2, a2p(ab)q = x2, a2r (ab)s = xy Assume π1(Mϕ) is left-ordered. Then th ...
... Proposition: Suppose p, q are non-negative and r, s are nonpositive integers (or vice-versa). Then π1(Mϕ) is not left-orderable. Proof: The group π1(Mϕ) has presentation with generators a, b, x, y and relations: a2 = b2, x2 = y 2, a2p(ab)q = x2, a2r (ab)s = xy Assume π1(Mϕ) is left-ordered. Then th ...
4. Lecture 4 Visualizing rings We describe several ways - b
... of X (the maximal ideal of a point is the ideal of functions vanishing there). The topology can be recovered as follows: take any ideal I of R, and let Z(I) be the set of maximal ideals containing I. These give the closed sets of X. (Different ideals can give the same closed set.) We can now do the ...
... of X (the maximal ideal of a point is the ideal of functions vanishing there). The topology can be recovered as follows: take any ideal I of R, and let Z(I) be the set of maximal ideals containing I. These give the closed sets of X. (Different ideals can give the same closed set.) We can now do the ...
4.3 Existence of Roots
... has order 3, so it is not a generator. Since 1 + x ≡ −x2 (mod x2 + x + 1), we have [1 + x]2 = [x] and [1 + x]3 = −[x]6 = −([x]3 )2 = −[1]. Thus [1 + x]6 = [1], but no smaller power is the identity, showing that [1 + x] has order 6, so it is not a generator. ...
... has order 3, so it is not a generator. Since 1 + x ≡ −x2 (mod x2 + x + 1), we have [1 + x]2 = [x] and [1 + x]3 = −[x]6 = −([x]3 )2 = −[1]. Thus [1 + x]6 = [1], but no smaller power is the identity, showing that [1 + x] has order 6, so it is not a generator. ...
Facts about finite fields
... Theoretically, the representation of a finite field makes no difference; however, when doing computations choosing a good “defining polynomial” f (x) can make a huge difference in running time, especially when using fields of large extension degree. Field containment and algebraic closure. We can co ...
... Theoretically, the representation of a finite field makes no difference; however, when doing computations choosing a good “defining polynomial” f (x) can make a huge difference in running time, especially when using fields of large extension degree. Field containment and algebraic closure. We can co ...
18. Cyclotomic polynomials II
... There are other possibly more natural approaches as well, but the following trick is worth noting. The ij th entry of V is ζ (i−1)(j−1) . Thus, the ij th entry of the square V 2 is ...
... There are other possibly more natural approaches as well, but the following trick is worth noting. The ij th entry of V is ζ (i−1)(j−1) . Thus, the ij th entry of the square V 2 is ...
m\\*b £«**,*( I) kl)
... Barnes [l] has constructed an example of a commutative semisimple normed annihilator algebra which is not a dual algebra. His example is not complete and when completed acquires a nonzero radical. In this paper we construct an example which is complete. The theory of annihilator algebras is develope ...
... Barnes [l] has constructed an example of a commutative semisimple normed annihilator algebra which is not a dual algebra. His example is not complete and when completed acquires a nonzero radical. In this paper we construct an example which is complete. The theory of annihilator algebras is develope ...
Practice Exam 1
... (b) List the cosets of H in G. (Since G is abelian, left and right cosets are the same.) [3] Let G be the group of 2 by 2 matrices whose entries are integers mod 7, and whose determinant is nonzero mod 7. Let H be the subset of G consisting of all matrices whose determinant is 1 mod 7. (a) How many ...
... (b) List the cosets of H in G. (Since G is abelian, left and right cosets are the same.) [3] Let G be the group of 2 by 2 matrices whose entries are integers mod 7, and whose determinant is nonzero mod 7. Let H be the subset of G consisting of all matrices whose determinant is 1 mod 7. (a) How many ...
Math 110 Homework 9 Solutions
... Note that the two polynomials in your example are still considered distinct elements in Fp [x], even if they define the same function Fp → Fp . Solution: (a) An irreducible polynomial is a polynomial such that if P (x) = q(x)r(x) then either q(x) or r(x) is a unit. (A unit is an element with a multi ...
... Note that the two polynomials in your example are still considered distinct elements in Fp [x], even if they define the same function Fp → Fp . Solution: (a) An irreducible polynomial is a polynomial such that if P (x) = q(x)r(x) then either q(x) or r(x) is a unit. (A unit is an element with a multi ...
p-Groups - Brandeis
... A p-group is a finite group P of order pk where k ≥ 0. Note that every subgroup of a p-group is a p-group. When we want to exclude the trivial case k = 0 we say that P is a nontrivial p-group. One of the most important properties of p-groups is that they have nontrivial centers: Theorem 3.1. Every n ...
... A p-group is a finite group P of order pk where k ≥ 0. Note that every subgroup of a p-group is a p-group. When we want to exclude the trivial case k = 0 we say that P is a nontrivial p-group. One of the most important properties of p-groups is that they have nontrivial centers: Theorem 3.1. Every n ...
Solutions — Ark 1
... Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible. (Hint: Use that k[X1 , . . . , Xn ] is UFD.) Solution: In fact, we are going to show that in any ring A being a UFD ...
... Oppgave 1. Show that the principal ideal (P (X1 , . . . , Xn )) in the polynomial ring k[X1 , . . . , Xn ] over the field k is prime if and only if P (X1 , . . . , Xn ) is irreducible. (Hint: Use that k[X1 , . . . , Xn ] is UFD.) Solution: In fact, we are going to show that in any ring A being a UFD ...
Algebra_Aug_2008
... 5) a) Solution 1: (Without galois theory) We can assume without loss of generality that f is monic, so f (x) is the minimal polynomial of α over k. But by assumption, α is also a root of f (x + 1), so f (x)|f (x + 1), hence f (x) = f (x + 1) since they are both monic of the same degree. Therefore α ...
... 5) a) Solution 1: (Without galois theory) We can assume without loss of generality that f is monic, so f (x) is the minimal polynomial of α over k. But by assumption, α is also a root of f (x + 1), so f (x)|f (x + 1), hence f (x) = f (x + 1) since they are both monic of the same degree. Therefore α ...
Algebraic closure
... S = {(k, a0 , a1 , a2 , · · · , an , 0, 0, · · ·) ∈ N × F × F × · · · | ai ∈ F, 1 6 k 6 n}. Any algebraic field extension E of F can have at most as many elements as the set S. (Every α ∈ E is a root of some polynomial f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ F [x], which has at most n different ...
... S = {(k, a0 , a1 , a2 , · · · , an , 0, 0, · · ·) ∈ N × F × F × · · · | ai ∈ F, 1 6 k 6 n}. Any algebraic field extension E of F can have at most as many elements as the set S. (Every α ∈ E is a root of some polynomial f (x) = a0 + a1 x + a2 x2 + · · · + an xn ∈ F [x], which has at most n different ...
Exercises MAT2200 spring 2013 — Ark 9 Field extensions and
... Ark9: Field extensions and Unique factorisation MAT2200 — Vår 2013 Oppgave 10. ( Section 31, No.: 30 on page 292 in the book). Let F ✓ E be a field extension. Assume that ↵ 2 E is algebraic over F of odd degree. Show that ↵2 is algebraic of odd degree and that F (↵) = F (↵2 ). Oppgave 11. Show that ...
... Ark9: Field extensions and Unique factorisation MAT2200 — Vår 2013 Oppgave 10. ( Section 31, No.: 30 on page 292 in the book). Let F ✓ E be a field extension. Assume that ↵ 2 E is algebraic over F of odd degree. Show that ↵2 is algebraic of odd degree and that F (↵) = F (↵2 ). Oppgave 11. Show that ...
Math 594, HW7
... c). Given A = (ca , da ), B = (cb , db ) the line equation is (cb −ca )(y−da ) = (db −da )(x−ca ) , and the circle equation is (WLOG - the center is at A) (x−ca )2 +(y−da )2 = (cb −ca )2 +(db −da )2 d). Intersection (x0 , y0 ) of 0 = ax + by + c and 0 = dx + ey + f , if exists, is the solution of a ...
... c). Given A = (ca , da ), B = (cb , db ) the line equation is (cb −ca )(y−da ) = (db −da )(x−ca ) , and the circle equation is (WLOG - the center is at A) (x−ca )2 +(y−da )2 = (cb −ca )2 +(db −da )2 d). Intersection (x0 , y0 ) of 0 = ax + by + c and 0 = dx + ey + f , if exists, is the solution of a ...
test solutions 2
... generated, so M = hm1 , . . . , mn i. We will prove the claim by induction on n. If n = 1, then M ∼ = Z/(a) for some a ∈ Z. If a = 0 then M ∼ = Z which is not Artinian. Otherwise a 6= 0 when M is finite. As quotients and submodules of Artinian modules are Artinian, in the general case N = hm1 , . . ...
... generated, so M = hm1 , . . . , mn i. We will prove the claim by induction on n. If n = 1, then M ∼ = Z/(a) for some a ∈ Z. If a = 0 then M ∼ = Z which is not Artinian. Otherwise a 6= 0 when M is finite. As quotients and submodules of Artinian modules are Artinian, in the general case N = hm1 , . . ...
Why is addition of fractions defined the way it is? Two reasons
... The additive identity is 0/1 The multiplicative idenity is 1/1 The additive inverse of a/b is (−a)/b The multiplicative inverse of a/b (when a 6= 0) is b/a. In this way, every integral domain can be embedded inside a field. Examples: Starting with Z we get Q. Starting with R[x] we get the rational f ...
... The additive identity is 0/1 The multiplicative idenity is 1/1 The additive inverse of a/b is (−a)/b The multiplicative inverse of a/b (when a 6= 0) is b/a. In this way, every integral domain can be embedded inside a field. Examples: Starting with Z we get Q. Starting with R[x] we get the rational f ...
Review Problems
... ac + b + d = −2, ad + bc = 8, and bd = 1. We have either b = d = 1, in which case a + c = 8, or b = d = −1, in which case a + c = −8. Either case contradicts a + c = 0, so x4 − 2x2 + 8x + 1 is irreducible over Q. Alternate solution: Once we have the factorization x5 − 2x4 − 2x3 + 12x2 − 15x − 2 = (x ...
... ac + b + d = −2, ad + bc = 8, and bd = 1. We have either b = d = 1, in which case a + c = 8, or b = d = −1, in which case a + c = −8. Either case contradicts a + c = 0, so x4 − 2x2 + 8x + 1 is irreducible over Q. Alternate solution: Once we have the factorization x5 − 2x4 − 2x3 + 12x2 − 15x − 2 = (x ...
aa2.pdf
... ) = 1. Use the Chinese remainder • Approach 2: Let f ∈ k[x] be such that gcd(f, dx theorem to construct inductively polynomials gr ∈ k[x], r = 1, 2, . . . , such that, setting pr := x + f ·g1 + f 2 ·g2 + . . . + f r ·gr ∈ k[x], we have f (pr (x)) ∈ (f (x))r · k[x]. Deduce in particular that, for any ...
... ) = 1. Use the Chinese remainder • Approach 2: Let f ∈ k[x] be such that gcd(f, dx theorem to construct inductively polynomials gr ∈ k[x], r = 1, 2, . . . , such that, setting pr := x + f ·g1 + f 2 ·g2 + . . . + f r ·gr ∈ k[x], we have f (pr (x)) ∈ (f (x))r · k[x]. Deduce in particular that, for any ...
Math 400 Spring 2016 – Test 3 (Take
... Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Prove each subgroup is normal in the one larger by noting index 2; prove K 6C G by noting dvd 6∈ K. 2. (8) Let G be the quaternion group (also known as Q8 ) that you worked with on Tes ...
... Solution: It’s false. The smallest (lowest order) counterexample is G = D4 , H = {R0 , R180 , v, h}, K = {R0 , v}. (Prove each subgroup is normal in the one larger by noting index 2; prove K 6C G by noting dvd 6∈ K. 2. (8) Let G be the quaternion group (also known as Q8 ) that you worked with on Tes ...
4.2 Factors - NIU Math Department
... Over Z5 , check that there are no roots. For the factorization, we get (x2 −2)(x2 +2) = x4 − 4 ≡ x4 + 1. Over Z7 , there are no roots. We get (x2 + 3x + 1)(x2 − 3x + 1) = x4 − 7x2 + 1 ≡ x4 + 1. Over Z11 , there are no roots. We get (x2 +3x−1)(x2 −3x−1) = x4 −11x2 +1 ≡ x4 +1. Note that since x4 + 1 h ...
... Over Z5 , check that there are no roots. For the factorization, we get (x2 −2)(x2 +2) = x4 − 4 ≡ x4 + 1. Over Z7 , there are no roots. We get (x2 + 3x + 1)(x2 − 3x + 1) = x4 − 7x2 + 1 ≡ x4 + 1. Over Z11 , there are no roots. We get (x2 +3x−1)(x2 −3x−1) = x4 −11x2 +1 ≡ x4 +1. Note that since x4 + 1 h ...
Full text
... In a previous paper [l] the idea of a "sieve" was extended in o r d e r to give a method for the computation of the sequences of values for certain functions which occur in the theory of numbers. ...
... In a previous paper [l] the idea of a "sieve" was extended in o r d e r to give a method for the computation of the sequences of values for certain functions which occur in the theory of numbers. ...
Wedderburn`s Theorem on Division Rings: A finite division ring is a
... (1) If V is a vector space of dimension n over a finite field F with |F | = q (note q ≥ 2, because any field contains both a 0 and a 1), then because V ∼ = F n as vector spaces, we have |V | = q n . In particular, if R is a finite ring containing a field F with q elements, then it is a vector space ...
... (1) If V is a vector space of dimension n over a finite field F with |F | = q (note q ≥ 2, because any field contains both a 0 and a 1), then because V ∼ = F n as vector spaces, we have |V | = q n . In particular, if R is a finite ring containing a field F with q elements, then it is a vector space ...
non-abelian classfields over function fields in special cases
... defined below, and tempted us to propose a following series of conjectures (C 1) ~ (C 5). It is stronger than the classfield theory of T-type defined in § 1.2. So far, it is proved only for some special cases of F, but seems to be supported also by some other results on T (e. g., § 3.1), We shall sp ...
... defined below, and tempted us to propose a following series of conjectures (C 1) ~ (C 5). It is stronger than the classfield theory of T-type defined in § 1.2. So far, it is proved only for some special cases of F, but seems to be supported also by some other results on T (e. g., § 3.1), We shall sp ...
Model Solutions
... under addition, multiplication and additive inverse, and include the constantly 0 function. The constantly 0 function is describable by the 0 polynomial. The sum f + g is describable by the sum of polynomials describing f and g; the additive inverse of f is describable by the additive inverse of a p ...
... under addition, multiplication and additive inverse, and include the constantly 0 function. The constantly 0 function is describable by the 0 polynomial. The sum f + g is describable by the sum of polynomials describing f and g; the additive inverse of f is describable by the additive inverse of a p ...
Exercises 01 [1.1]
... That is, given an abelian group Y and group homomorphisms fn : Y → Z compatible with the transition maps (meaning that 2 · fn (y) = fn−1 (y) for y ∈ Y ), there is a unique f : Y → X making a commutative diagram ...
... That is, given an abelian group Y and group homomorphisms fn : Y → Z compatible with the transition maps (meaning that 2 · fn (y) = fn−1 (y) for y ∈ Y ), there is a unique f : Y → X making a commutative diagram ...